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Add related-work section, refute floor-containment conjecture

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Introduction now positions the tire-tree decomposition against
Birkhoff, Tutte, Heesch, Robertson-Sanders-Seymour-Thomas, and
Dvorak-Lidicky's coloring count cones (closest modern parallel).

Floor-containment conjecture refuted at n=4 and n=6: explicit
counterexample colorings (1,2,1,2), (1,3,2,1,3,2), (2,3,2,3,2,3)
absent from non-floor supports.  Skip-m=3 sweep through m=8 partial
still consistent with floor-stability-in-m.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
didericis
2026-06-03 00:09:58 -04:00
parent 57f5c2839a
commit 35d226f8f8
7 changed files with 288 additions and 154 deletions
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papers/coloring_nested_tire_graphs/experiments/level_cycle_support.py
+33
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@@ -327,6 +327,38 @@ def enumerate_level_cycle_supports(
return supports, stats return supports, stats
def check_floor_containment(
supports: dict[frozenset[tuple[int, ...]], list[dict]],
) -> None:
"""Test whether the smallest-size support is a subset of every other support.
The supports are fully normalized (orbits under S_4 x D_n), so set
inclusion here is the right test for the "floor support is contained in
every admissible support" conjecture.
"""
if len(supports) < 2:
print(" containment: <2 supports, trivially contained")
return
keys = list(supports.keys())
min_size = min(len(k) for k in keys)
floors = [k for k in keys if len(k) == min_size]
print(f" containment check: floor support (size {min_size}) vs {len(keys) - 1} others")
for floor in floors:
bad: list[tuple[int, tuple[int, ...]]] = []
for other in keys:
if other is floor:
continue
if not floor.issubset(other):
missing = next(iter(floor - other))
bad.append((len(other), missing))
if not bad:
print(f" floor IS a subset of all {len(keys) - 1} other supports")
else:
print(f" floor is NOT a subset of {len(bad)} of {len(keys) - 1} other supports")
for size, missing in bad[:5]:
print(f" missing example: floor coloring {missing} absent from support of size {size}")
def summarize( def summarize(
n: int, n: int,
supports: dict[frozenset[tuple[int, ...]], list[dict]], supports: dict[frozenset[tuple[int, ...]], list[dict]],
@@ -357,6 +389,7 @@ def summarize(
print(f" most restrictive support : {min_size}/{all_colorings} ({len(min_supports)} support types)") print(f" most restrictive support : {min_size}/{all_colorings} ({len(min_supports)} support types)")
print(f" least restrictive support : {max_size}/{all_colorings}") print(f" least restrictive support : {max_size}/{all_colorings}")
print(f" overlap among max-restrict : min {min(pair_overlaps)}, max {max(pair_overlaps)}") print(f" overlap among max-restrict : min {min(pair_overlaps)}, max {max(pair_overlaps)}")
check_floor_containment(supports)
print(" examples:") print(" examples:")
printed = 0 printed = 0
papers/coloring_nested_tire_graphs/experiments/level_cycle_support_findings.md
+70
View File
@@ -239,3 +239,73 @@ opens the door to closed-form support computation for the worst case.
The obstruction to a quick proof is exotic `m ≥ 4` configurations not The obstruction to a quick proof is exotic `m ≥ 4` configurations not
reachable by subdivision from an `m = 3` floor witness — none has reachable by subdivision from an `m = 3` floor witness — none has
appeared empirically, but their non-existence is not yet ruled out. appeared empirically, but their non-existence is not yet ruled out.
#### Skip-`m=3` sweep (partial)
To probe the conjecture more directly we ran a "skip-`m=3`" sweep
explicitly excluding `m = 3` from the search:
```bash
python3 -u papers/coloring_nested_tire_graphs/experiments/level_cycle_support.py \
--n-min 6 --n-max 6 --outer-min 4 --outer-max 10 --inner-min 7 --inner-max 9 \
--progress
```
The sweep ran for `11h 49m` and was terminated partway through `m = 8,
k = 9` at chord-set #6 of the 2-chord block (raw=6.45M canonical=223k).
Throughout the run --- across `m ∈ [4, 7]` exhaustively and `m = 8`
partially --- the distinct-support count stayed locked at **`13`**
(vs `19` for the full `m ∈ [3, 7]` run). The six missing support
types are precisely the ones with `m = 3`-only witnesses, including
the unique `252/732` floor.
Conclusion of the partial run: no support of size `≤ 252` appeared
under `m ≥ 4` in any configuration searched, and the support count
saturated early, suggesting the support landscape for `m ≥ 4` is
strictly looser than for `m = 3` even as `m` grows. The conjecture is
not falsified by any data we have; pushing `m` to `9` or `10` is
left as future work (path-count growth makes it expensive: `m = 10,
k = 9` alone has `C(19, 10) ≈ 92{,}000` paths per chord set).
### Floor-containment conjecture — refuted
Floor-stability bounds the *size* of the worst-case support but not
its content. For the tire-tree compatibility argument, what would
have closed the loop is a containment statement:
> **(Conjecture, refuted)** For every `n`, the unique floor support
> `F_n` is a *subset* of every admissible level-cycle support. Every
> level-cycle coloring that the `m = 3` floor witness allows is also
> allowed by every other tire.
The `check_floor_containment` helper in `level_cycle_support.py`
tests this directly on the computed supports. The conjecture **fails
at `n = 4` and `n = 6`**:
```text
n=4: floor (size 60) is NOT a subset of 1 of 2 other supports
missing: floor coloring (1, 2, 1, 2) absent from support of size 72
n=6: floor (size 252) is NOT a subset of 11 of 16 other supports
missing: floor coloring (1, 3, 2, 1, 3, 2) absent from support of size 708
missing: floor coloring (2, 3, 2, 1, 3, 1) absent from support of size 492
missing: floor coloring (2, 3, 2, 3, 2, 3) absent from support of size 720
```
(`n = 3` is trivial — only one support type. `n = 5` happens to
satisfy containment in the searched window but only against a single
other support, so the data is too thin to be evidence either way.)
The missing colorings cluster on **bipartite alternations**
(`(1,2,1,2)`, `(2,3,2,3,2,3)`) and **3-periodic patterns**
(`(1,3,2,1,3,2)`). These are the most "structured" colorings of `C_n`
— colorings the thin `m = 3` annulus is free to produce, but other
tires (with larger annulus / heavier chord patterns) actively forbid
because their interior forces additional colour variety.
**Consequence.** The simple structural shortcut — "all supports
contain a common floor, hence pairwise intersections are trivially
non-empty" — does not hold. Same-`n` compatibility (which by `4CT` is
still forced to hold) cannot be reduced to a single containment fact;
it requires a finer structural analysis of *which* colorings sit in
*which* supports, or a different shortcut entirely.
papers/coloring_nested_tire_graphs/paper.aux
+43 -26
View File
@@ -1,72 +1,89 @@
\relax \relax
\citation{bauerfeld-nested-tire-duals} \citation{bauerfeld-nested-tire-duals}
\citation{birkhoff-reducibility}
\citation{birkhoff-lewis-chromatic}
\citation{tutte-chromatic-sums-1973}
\citation{tutte-algebraic-colorings}
\citation{tutte-four-colour-conjecture}
\citation{dvorak-lidicky-cones}
\citation{heesch-untersuchungen}
\citation{robertson-sanders-seymour-thomas}
\citation{robertson-sanders-seymour-thomas}
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\@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces A tire graph with non-degenerate boundaries: outer boundary $B_{\mathrm {out}}$ a $6$-cycle on vertices $0,\dots ,5$ (blue), inner boundary $B_{\mathrm {in}}$ a $4$-cycle on vertices $6,\dots ,9$ (red), inner outerplanar graph $O = B_{\mathrm {in}} \cup \{7\text {--}9\}$ (with one chord, orange), and $E_{\mathrm {ann}}$ (grey) tiling the annulus between $B_{\mathrm {out}}$ and $B_{\mathrm {in}}$ by ten triangular faces.}}{4}{}\protected@file@percent } \@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces A tire graph with non-degenerate boundaries: outer boundary $B_{\mathrm {out}}$ a $6$-cycle on vertices $0,\dots ,5$ (blue), inner boundary $B_{\mathrm {in}}$ a $4$-cycle on vertices $6,\dots ,9$ (red), inner outerplanar graph $O = B_{\mathrm {in}} \cup \{7\text {--}9\}$ (with one chord, orange), and $E_{\mathrm {ann}}$ (grey) tiling the annulus between $B_{\mathrm {out}}$ and $B_{\mathrm {in}}$ by ten triangular faces.}}{4}{}\protected@file@percent }
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\citation{bauerfeld-depth} \citation{bauerfeld-depth}
\newlabel{lem:tire-component}{{1.9}{5}} \newlabel{lem:tire-component}{{1.9}{5}}
\citation{bauerfeld-depth} \citation{bauerfeld-depth}
\newlabel{thm:tread-partition}{{1.10}{6}} \newlabel{thm:tread-partition}{{1.10}{6}}
\newlabel{rem:tire-component-degenerate}{{1.11}{7}} \newlabel{rem:tire-component-degenerate}{{1.11}{7}}
\newlabel{rem:tire-no-extra-hypotheses}{{1.12}{7}} \newlabel{rem:tire-no-extra-hypotheses}{{1.12}{7}}
\newlabel{thm:inner-dual-outerplanar}{{1.13}{7}} \newlabel{thm:inner-dual-outerplanar}{{1.13}{8}}
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\newlabel{fig:inner-dual-disk-case}{{3}{8}} \newlabel{fig:inner-dual-disk-case}{{3}{8}}
\citation{bauerfeld-nested-tire-duals} \citation{bauerfeld-nested-tire-duals}
\citation{bauerfeld-nested-tire-duals} \citation{bauerfeld-nested-tire-duals}
\citation{tait-original}
\newlabel{rem:hamilton-cycle-spoke-only}{{1.14}{9}} \newlabel{rem:hamilton-cycle-spoke-only}{{1.14}{9}}
\newlabel{rem:bridge-case-theta}{{1.15}{9}} \newlabel{rem:bridge-case-theta}{{1.15}{9}}
\newlabel{thm:tait-tire}{{1.16}{9}} \citation{tait-original}
\@writefile{lof}{\contentsline {figure}{\numberline {4}{\ignorespaces Case 2 ($R$ = annulus) with $O$ a barbell. $B_{\mathrm {out}}$ is the outer hexagon (red); $O$ has two triangles $\{a_1, a_2, a_3\}$ and $\{b_1, b_2, b_3\}$ joined by the bridge $a_3\text {--}b_1$ (all light red). The annulus is triangulated by $14$ annular triangles: $6$ ``outer-cap'' triangles (one per outer edge), $6$ ``inner-cap'' triangles (one per non-bridge edge of $O$), and $2$ ``bridge-cap'' triangles $\{u_0, a_3, b_1\}$ and $\{u_3, a_3, b_1\}$ adjacent to the bridge. Each blue dot sits at the centroid of an annular triangle; blue edges connect dual vertices whose triangles share an interior annular edge (spoke or bridge). The two bridge-cap vertices have $\Gamma $-degree $3$ (their triangles have no boundary edge) and are joined by the dashed blue \emph {chord} corresponding to the bridge; the remaining $13$ edges form the Hamilton cycle that wraps around the annulus. All $14$ vertices lie on the outer face of the cycle-with-chord embedding, so $\Gamma \cong \Theta (1, 7, 7)$ is outerplanar.}}{10}{}\protected@file@percent } \@writefile{lof}{\contentsline {figure}{\numberline {4}{\ignorespaces Case 2 ($R$ = annulus) with $O$ a barbell. $B_{\mathrm {out}}$ is the outer hexagon (red); $O$ has two triangles $\{a_1, a_2, a_3\}$ and $\{b_1, b_2, b_3\}$ joined by the bridge $a_3\text {--}b_1$ (all light red). The annulus is triangulated by $14$ annular triangles: $6$ ``outer-cap'' triangles (one per outer edge), $6$ ``inner-cap'' triangles (one per non-bridge edge of $O$), and $2$ ``bridge-cap'' triangles $\{u_0, a_3, b_1\}$ and $\{u_3, a_3, b_1\}$ adjacent to the bridge. Each blue dot sits at the centroid of an annular triangle; blue edges connect dual vertices whose triangles share an interior annular edge (spoke or bridge). The two bridge-cap vertices have $\Gamma $-degree $3$ (their triangles have no boundary edge) and are joined by the dashed blue \emph {chord} corresponding to the bridge; the remaining $13$ edges form the Hamilton cycle that wraps around the annulus. All $14$ vertices lie on the outer face of the cycle-with-chord embedding, so $\Gamma \cong \Theta (1, 7, 7)$ is outerplanar.}}{10}{}\protected@file@percent }
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\newlabel{thm:tire-chromatic-polynomial-transfer}{{1.19}{11}} \newlabel{thm:tire-chromatic-polynomial-transfer}{{1.19}{12}}
\newlabel{rem:spoke-only-chromatic-transfer}{{1.20}{12}} \newlabel{rem:spoke-only-chromatic-transfer}{{1.20}{12}}
\newlabel{thm:tread-tree}{{1.21}{12}} \newlabel{thm:tread-tree}{{1.21}{13}}
\newlabel{rem:tree-multiple-children}{{1.22}{13}} \newlabel{rem:tree-multiple-children}{{1.22}{14}}
\newlabel{thm:tire-tree-decomposition}{{1.23}{13}} \newlabel{thm:tire-tree-decomposition}{{1.23}{14}}
\newlabel{rem:tree-coloring-factorisation}{{1.24}{15}}
\@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces Tire-tree decomposition (Theorem\nonbreakingspace 1.23\hbox {}) on a $13$-vertex maximal planar example $G$ with five BFS levels. $(a)$ $G$ with vertex source $v_0$ and $\ell _G \in \{0,1,2,3,4\}$; four nested seams are highlighted, $C_{T_R} = \{a,b,c\}$ (orange), $C_{T_L} = \{a,c,d\}$ (red, including the chord $a$-$c$ shared with $C_{T_R}$), $C_{T_{LL}} = \{f_1, f_2, f_3\}$ (purple), $C_{T_{LLL}} = \{g_1, g_2, g_3\}$ (teal). Inset: the rooted tree of tire treads $\mathcal {T}(G, \{v_0\})$ branches at $T_0$ into the leaf $T_R$ (containing $e$) and a chain $T_L \to T_{LL} \to T_{LLL}$ (the highlighted sub-tree). $(b)$ The disk $G_{T_L}$ inside the seam $C_{T_L}$, drawn standalone with $C_{T_L}$ as cycle source and vertex labels rotated to match the new (cycle-source) role of the boundary triangle. $\ell _{G_{T_L}}(\cdot ) = \ell _G(\cdot ) - 1$ on $V(G_{T_L})$ (verified by the generator script), and $\mathcal {T}(G_{T_L}, C_{T_L})$ is the chain $T_L \to T_{LL} \to T_{LLL}$, iso to the highlighted sub-tree of $(a)$.}}{16}{}\protected@file@percent } \@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces Tire-tree decomposition (Theorem\nonbreakingspace 1.23\hbox {}) on a $13$-vertex maximal planar example $G$ with five BFS levels. $(a)$ $G$ with vertex source $v_0$ and $\ell _G \in \{0,1,2,3,4\}$; four nested seams are highlighted, $C_{T_R} = \{a,b,c\}$ (orange), $C_{T_L} = \{a,c,d\}$ (red, including the chord $a$-$c$ shared with $C_{T_R}$), $C_{T_{LL}} = \{f_1, f_2, f_3\}$ (purple), $C_{T_{LLL}} = \{g_1, g_2, g_3\}$ (teal). Inset: the rooted tree of tire treads $\mathcal {T}(G, \{v_0\})$ branches at $T_0$ into the leaf $T_R$ (containing $e$) and a chain $T_L \to T_{LL} \to T_{LLL}$ (the highlighted sub-tree). $(b)$ The disk $G_{T_L}$ inside the seam $C_{T_L}$, drawn standalone with $C_{T_L}$ as cycle source and vertex labels rotated to match the new (cycle-source) role of the boundary triangle. $\ell _{G_{T_L}}(\cdot ) = \ell _G(\cdot ) - 1$ on $V(G_{T_L})$ (verified by the generator script), and $\mathcal {T}(G_{T_L}, C_{T_L})$ is the chain $T_L \to T_{LL} \to T_{LLL}$, iso to the highlighted sub-tree of $(a)$.}}{16}{}\protected@file@percent }
\newlabel{fig:tire-tree-decomposition}{{5}{16}} \newlabel{fig:tire-tree-decomposition}{{5}{16}}
\newlabel{rem:level-cycle-motivation}{{1.25}{16}} \newlabel{rem:tree-coloring-factorisation}{{1.24}{16}}
\newlabel{def:level-cycle-three-colour-restriction}{{1.26}{16}} \newlabel{rem:level-cycle-motivation}{{1.25}{17}}
\newlabel{def:level-cycle-three-colour-restriction}{{1.26}{17}}
\newlabel{conj:false-universal-level-cycle-three-colour}{{1.27}{17}} \newlabel{conj:false-universal-level-cycle-three-colour}{{1.27}{17}}
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\newlabel{tab:level-cycle-three-colour-c5-14-16}{{2}{21}} \newlabel{def:seam}{{1.33}{21}}
\newlabel{def:partial-tire-tree}{{1.35}{21}} \newlabel{def:partial-tire-tree}{{1.34}{21}}
\newlabel{lem:seam-edge-shared}{{1.36}{21}} \newlabel{lem:seam-edge-shared}{{1.35}{21}}
\newlabel{conj:seam-counterexample}{{1.37}{21}} \newlabel{conj:seam-counterexample}{{1.36}{21}}
\bibcite{tait-original}{1} \bibcite{tait-original}{1}
\bibcite{bauerfeld-depth}{2} \bibcite{bauerfeld-depth}{2}
\bibcite{bauerfeld-nested-tire-duals}{3} \bibcite{bauerfeld-nested-tire-duals}{3}
\bibcite{birkhoff-reducibility}{4}
\bibcite{birkhoff-lewis-chromatic}{5}
\bibcite{tutte-four-colour-conjecture}{6}
\bibcite{tutte-algebraic-colorings}{7}
\bibcite{tutte-chromatic-sums-1973}{8}
\bibcite{heesch-untersuchungen}{9}
\bibcite{robertson-sanders-seymour-thomas}{10}
\bibcite{dvorak-lidicky-cones}{11}
\newlabel{tocindent-1}{0pt} \newlabel{tocindent-1}{0pt}
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@@ -82,6 +82,40 @@ companion paper \cite{bauerfeld-nested-tire-duals} uses these
definitions to develop nested-cycle structure theorems and definitions to develop nested-cycle structure theorems and
chain-pigeonhole conjectures for tire annular subgraphs of $G'$. chain-pigeonhole conjectures for tire annular subgraphs of $G'$.
\paragraph{Related work.}
The structural object underlying this programme --- the set of
proper $4$-colourings of a boundary cycle that extend to a colouring
of a bounded planar region --- is classical. Birkhoff's reducibility
analysis of the diamond configuration~\cite{birkhoff-reducibility} is
the earliest instance of computing such extension sets to attack the
Four Colour Theorem; the chromatic polynomial framework of Birkhoff
and Lewis~\cite{birkhoff-lewis-chromatic} systematised the counting.
Tutte studied how the chromatic polynomial of a rooted planar
triangulation decomposes along its outer
boundary~\cite{tutte-chromatic-sums-1973} and developed an algebraic
theory of graph colourings organised around separating
subgraphs~\cite{tutte-algebraic-colorings, tutte-four-colour-conjecture}.
The most recent and structurally closest parallel is Dvo\v{r}\'ak
and Lidick\'y's analysis of \emph{coloring count
cones}~\cite{dvorak-lidicky-cones}, which characterises the possible
boundary-extension functions on a fixed outer cycle of a
near-triangulation. The Heesch--Appel--Haken
approach~\cite{heesch-untersuchungen, robertson-sanders-seymour-thomas}
also uses boundary-extension reasoning, but case-by-case on a finite
unavoidable set of local configurations rather than as part of a
global structural induction.
The tire-tree decomposition introduced here differs from each of
these in shape rather than ingredients. Birkhoff, Tutte, and
Dvo\v{r}\'ak--Lidick\'y all study \emph{one} boundary; Heesch and
the cleaned-up Appel--Haken proof~\cite{robertson-sanders-seymour-thomas}
study a finite collection of local boundaries. The present framework
organises the entire triangulation into a hierarchy of annular
regions glued along level cycles, and asks whether boundary-extension
constraints compose compatibly up the hierarchy. To the authors'
knowledge, no prior work on the Four Colour Theorem has been
organised around a global nested-cycle decomposition of this kind.
Throughout, $G = (V, E)$ is a plane maximal planar graph (a triangulation) Throughout, $G = (V, E)$ is a plane maximal planar graph (a triangulation)
with a fixed planar embedding $\Pi_G$. We write $|V| = n$, so $|E| = 3n - 6$ with a fixed planar embedding $\Pi_G$. We write $|V| = n$, so $|E| = 3n - 6$
and $G$ has $2n - 4$ triangular faces. and $G$ has $2n - 4$ triangular faces.
@@ -1373,35 +1407,6 @@ inner-boundary three-colour restriction with respect to
$\mathcal{T}(G, \{v_0\})$. $\mathcal{T}(G, \{v_0\})$.
\end{conjecture} \end{conjecture}
\begin{remark}[Relation to the level-cycle conjecture]
\label{rem:inner-boundary-vs-level-cycle}
For a depth-$d$ tire $T$, the inner outerplanar graph satisfies
$O^{(T)} \subseteq G[L_{d+1}]$: a depth-$d$ face has its three vertex
levels in $\{d, d+1\}$ (adjacent vertices differ by at most one level),
so the level-$(d+1)$ vertices of the tire's dual component are exactly
$V(O^{(T)})$. Since $O^{(T)}$ is outerplanar, every one of its vertices
lies on the inner-boundary walk, whence $V(B_{\mathrm{in}}^{(T)}) =
V(O^{(T)}) \subseteq L_{d+1}$ is supported on a single level, and is a
simple level cycle when $O^{(T)}$ is $2$-connected.
Consequently the vertex-source form of
Conjecture~\ref{conj:level-cycle-three-colour} implies
Conjecture~\ref{conj:tire-inner-boundary-three-colour} on every
$2$-connected inner boundary: the witnessing colouring already makes
each such cycle omit a colour. The present conjecture is thus a
\emph{weakening}, constraining only the inner-boundary cycles of one
tire-tree decomposition rather than all level cycles of some level
source. It is no harder than the vertex-source form of
Conjecture~\ref{conj:level-cycle-three-colour}, while targeting exactly
the interface the chromatic-transfer machinery of
Theorem~\ref{thm:tire-chromatic-polynomial-transfer} runs across.
(The non-$2$-connected case --- an
inner boundary whose walk traverses a bridge or cut-vertex of $O^{(T)}$
--- is not covered by the simple-cycle statement of
Conjecture~\ref{conj:level-cycle-three-colour} and must be argued
separately.)
\end{remark}
\subsection*{A counterexample at $n=14$} \subsection*{A counterexample at $n=14$}
Conjecture~\ref{conj:tire-inner-boundary-three-colour} is in fact Conjecture~\ref{conj:tire-inner-boundary-three-colour} is in fact
@@ -1452,38 +1457,18 @@ $17,28,69$ together with $12$.}
The failure was verified by enumerating, for each of the $14$ vertex The failure was verified by enumerating, for each of the $14$ vertex
sources, all $96$ proper $4$-colourings of $G^\star$ and computing the sources, all $96$ proper $4$-colourings of $G^\star$ and computing the
inner boundary $V(B_{\mathrm{in}}^{(T)})$ of every tire $T$ as the inner boundary $V(B_{\mathrm{in}}^{(T)})$ of every tire $T$ as the
level-$(d+1)$ vertices of the corresponding depth-$d$ dual component level-$(d+1)$ vertices of the corresponding depth-$d$ dual component.
(Remark~\ref{rem:inner-boundary-vs-level-cycle}). Each source has Each source has exactly two non-degenerate inner boundaries
exactly two non-degenerate inner boundaries (size $\geq 4$), and every (size $\geq 4$), and every proper $4$-colouring assigns all four
proper $4$-colouring assigns all four colours to at least one of them. colours to at least one of them.
Because Conjecture~\ref{conj:tire-inner-boundary-three-colour} is a The graph $G^\star$ does not refute
weakening of the vertex-source form of Conjecture~\ref{conj:level-cycle-three-colour}: the vertex source
Conjecture~\ref{conj:level-cycle-three-colour} only \emph{on $v_0 = 10$ admits a proper $4$-colouring under which every simple level
$2$-connected inner boundaries} cycle uses at most three colours.
(Remark~\ref{rem:inner-boundary-vs-level-cycle}), $G^\star$ need not
refute Conjecture~\ref{conj:level-cycle-three-colour}, and in fact does
not: the vertex source $v_0 = 10$ admits a proper $4$-colouring under
which every simple level cycle uses at most three colours. Combining
this with Remark~\ref{rem:inner-boundary-vs-level-cycle}, at least one
tire $T$ under $v_0 = 10$ must have an inner outerplanar graph
$O^{(T)}$ that fails to be $2$-connected --- under the witnessing
colouring, some inner boundary uses all four colours, and if its
$O^{(T)}$ were $2$-connected then by
Remark~\ref{rem:inner-boundary-vs-level-cycle} that inner boundary
would be a simple level cycle, contradicting the level-cycle witness.
The failure of
Conjecture~\ref{conj:tire-inner-boundary-three-colour} is therefore
attributable to the non-$2$-connected case left open by
Remark~\ref{rem:inner-boundary-vs-level-cycle}, not to a deeper failure
of the level-cycle statement on $G^\star$.
\subsection*{The surviving level-cycle conjecture} \subsection*{The surviving level-cycle conjecture}
The verification on $G^\star$ above is consistent with the broader
empirical picture for the level-cycle restriction, which we record as
the conjecture this section ultimately stands on.
\begin{conjecture}[Level-cycle three-colour conjecture] \begin{conjecture}[Level-cycle three-colour conjecture]
\label{conj:level-cycle-three-colour} \label{conj:level-cycle-three-colour}
Let $G$ be a maximal planar graph. Then there exists a level source Let $G$ be a maximal planar graph. Then there exists a level source
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\end{thebibliography} \end{thebibliography}
\end{document} \end{document}